It’s well known that (lunar) tides on Earth result in a transfer of angular momentum from Earth proper to the Earth–Moon orbital motion. That’s why the Moon resides now in a high Earth orbit, and Earth’s axial rotation is nowadays much slower than in the past (Hadean, Archean etc.).

What about effects of solar tides, indeed? The angular momentum certainly must be transferred from Earth proper to the Earth–Sun orbital motion. Does it result in a tiny increase of (Earth+Moon)’s heliocentric orbital angular momentum over millenia? What about Mars and Ceres: does the solar tide brake their axial rotation and change slightly their orbits? And what about Venus: does its solar tidal braking explain the slowness of rotation completely?

Addendum. The solar tide may not be confused with “dual” effect: planetary tide on the central star. It exists as well and has much significance for (extrasolar) planets on near orbits — but this question is not about tides on Sun. Ī̲’d accept an answer assuming gravity of the Sun to be spherically symmetric, neglecting Sun’s deviation from spherical symmetry, as well as rotation (even going to account for General Relativity). The solar tide is interaction between this gravity and the orbiting object. By Newton’s laws, gravitation affects Sun’s momentum reciprocally (and, hence, velocity), and we may ignore here any effect on Sun beyond its translational motion (momentum/velocity/position).

  • $\begingroup$ A small point, not wanting to give a full answer as my math gets rough, but the tidal effect of the planets on the sun is quite tiny because the planet's distance from the sun is significant compared to their size. Tiny dots in the sky don't create measurable tides and from the sun, planets are just little dots. There's some effect, but it's small, much smaller than the Moon's on the Earth. In fact, I (think) the sun shedding mass may be a bigger factor on the Earth's movement away from the sun than the tidal transfer of energy to orbital angular momentum. $\endgroup$
    – userLTK
    Sep 10 '16 at 2:24
  • $\begingroup$ @userLTK: The solar tide is an interaction between (inhomogeneous, but spherically symmetric) gravity of the Sun and the orbiting object. This object’s geometry and rotational motion is significant, not Sun’s. We are not going to discuss effects on the central body – it could be even a black hole, Ī̲ don’t care. $\endgroup$ Sep 10 '16 at 10:23
  • $\begingroup$ If the Sun was replaced by a black hole of the same mass, the transfer of energy to angular momentum would be zero because there would be no tidal bulge on the sun. The sun's size is a factor too. As it expands in size, the effect will increase. (unless I'm not understanding your question). The effect on Venus' slowing would be the same, but the suns size matters in the transfer of energy to the planet's orbit. $\endgroup$
    – userLTK
    Sep 10 '16 at 14:33
  • $\begingroup$ @userLTK: If the Sun was replaced by a Schwarzschild black hole of the same mass, the transfer of angular momentum from rotating planet would happen because there would be a tidal bulge on the planet. Try to draw a simple (Newtonian) force diagram assuming two point masses approximation for the planet. Still not understanding my question? Ī̲’m sorry, but Ī̲ applied due efforts to help. $\endgroup$ Sep 10 '16 at 15:26
  • $\begingroup$ @userLTK: You can better understand the mechanism if consider the central body orbiting about the centre of mass, not assuming it has an infinite mass. $\endgroup$ Sep 10 '16 at 15:58

You're right; the effect of solar tides on the Earth would be to increase the Earth's (or Earth+Moon's) orbital angular momentum around the Sun. But it would be difficult to compute this properly, and probably impossible to measure the effect, because of all the other small alterations to the Earth's orbit going on.

A simple way to see the effect (which I gather you've already worked out) is to imagine two equal-mass objects in circular orbit about their common center of mass (midway between the two), one effectively a point mass (e.g., black hole) and the other being the Earth. If all motion is prograde (Earth rotates in same direction as orbit) and the Earth rotates faster than the orbital period, then the Earth's tidal bulge will be slightly ahead of the Earth-Sun direction. Then the near-side bulge will experience a retarding force from the point mass (slowing the Earth's rotation) while simultaneously pulling on the point mass in approximately the same direction as the latter's orbital motion. So the point mass is accelerated, adding to the angular momentum and making the orbit larger.

I believe that most attempts to estimate the solar tidal torque on the Earth end up with it begin about 5 to 10 times weaker than the lunar torque, so most of the slowdown in the Earth's rotation is still due to the Moon.

Of course, in the real Solar System the Sun also experiences a tidal bulge raised by the Earth, which (because the Sun rotates faster than the Earth's orbit) will accelerate the Earth forward in its orbit, also adding to the orbital angular momentum. But then you have remember that Venus and Mercury will raise larger tidal bulges on the Sun with different periods (relative to the Earth's orbit), and so the combined effect gets complicated.

As for Mars and Ceres: yes, in principle the same effect would apply. But it would be weaker for Mars (smaller planetary radius, larger distance from Sun) and much weaker for Ceres (same reasons, except moreso).

In addition, all of the orbits are being affected by N-body interactions with the rest of the planets (especially Jupiter), which I strongly suspect means that any potential changes in orbital angular momentum due to solar tides would be washed out by the various N-body and resonant interactions.

(My understanding is that solar tidal braking could explain the current rotations of both Mercury and Venus, though you really can't be sure because you don't know what their initial rotations were.)

  • $\begingroup$ Ī̲ made some estimates for Mars. All its axial angular momentum is worth altering the orbit by ≈ 13 km (of which 5 km perpendicularly to the orbital plane), two orders of magnitude under that would turn the orbital plane by only 1″. You are right — effects of the solar tide on orbits are insignificant. $\endgroup$ Sep 13 '16 at 21:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.